# Thread: Determinants of this matrix - pls help

1. ## Determinants of this matrix - pls help

Hi, can someone help me out with this, how do I solve this matrix question:

Let A = [a,b,c; d,e,f; g,h,i] Given that detA= alpha and alpha is not equal to zero, express each of the following terms of alpha:

a) det(2A)

b) det(A^-1)

c) det [2a,2b,2c; d,e,f; g,h,i]

d) det [a,b,c; d,e,f; g,h,i]

e) det [a,b,c; d-3a, e-3b, f-3c; g,h,i]

Thank you

2. ## Re: Determinants of this matrix - pls help

Did you mean det(A)= alpha

3. ## Re: Determinants of this matrix - pls help

Yes, sorry that was a typo - I have corrected it. Thanks

4. ## Re: Determinants of this matrix - pls help

det(A) = a(ei - fh) - b(di-fg) + c(dh-eg) = alpha

Now:

det(2A) = 8a(ei - fh) - 8b(di-fg) + 8c(dh-eg) = 8(a(ei - fh) - b(di-fg) + c(dh-eg))= 8alpha

try the rest...

check Determinant - Wikipedia, the free encyclopedia

5. ## Re: Determinants of this matrix - pls help

Thank you Goku,

I have these answers, are these correct?

a) 8alpha
b) 1/alpha
c) 2alpha
d) alpha
e) alpha

6. ## Re: Determinants of this matrix - pls help

Hello, zzizi!

These are easy if you know the properties of determinants.

Let $\displaystyle A = \begin{bmatrix}a&b&c \\ d&e&f \\ g&h&i \end{bmatrix}$

Given that $\displaystyle |A| = \alpha\text{ and }\alpha \ne 0$, express each of the following terms of $\displaystyle \alpha:$

$\displaystyle (a)\;|2A|$

Answer: .$\displaystyle 8\alpha$

$\displaystyle (b)\;|A^{-1}|$

Answer: .$\displaystyle \frac{1}{\alpha}$

$\displaystyle (c)\;\begin{vmatrix}2a&2b&2c \\ d&e&f \\ g&h&i\end{vmatrix}\right|$

Answer: .$\displaystyle 2\alpha$

$\displaystyle (d)\;\begin{vmatrix}a&b&c \\ d&e&f \\ g&h&i\end{vmatrix}$

Answer: .$\displaystyle \alpha$

$\displaystyle (e)\;\begin{vmatrix}a&b&c \\ d-3a & e-3b & f-3c \\ g&h&i\end{vmatrix}$

Answer: .$\displaystyle \alpha$

7. ## Re: Determinants of this matrix - pls help

Thank you very much for your clear answer. How did you type it up like that?

8. ## Re: Determinants of this matrix - pls help

Im a little stuck with this one, how do I work out this determinant? can anyone help with this?

Thanks

$\displaystyle \begin{vmatrix}a &b &c \\d-a&e-b &f-c \\g-2a&h-2b &i-2c\end{vmatrix}$

9. ## Re: Determinants of this matrix - pls help

Use the property det(AB) = det(A).det(B)...

AB = $\displaystyle \begin{vmatrix}a &b &c \\d-a&e-b &f-c \\g-2a&h-2b &i-2c\end{vmatrix}$

Let B = $\displaystyle \begin{vmatrix}a &b &c \\d&e &f \\g&h &i\end{vmatrix}$

we already know det(B) = alpha

Now find a matrix A multiplied by B,that will give you AB...

I will leave this to you to do, if you have trouble finding sucha matrix then post again.
Then Find det(A) multiply by det(B)...

I have not done this in a long time, so please excuse if there are any mistakes.

10. ## Re: Determinants of this matrix - pls help

Hello, zzizi!

I'm a little stuck with this one. .How do I work out this determinant?

. . $\displaystyle \begin{vmatrix}a &b &c \\d-a&e-b &f-c \\g-2a&h-2b &i-2c\end{vmatrix}$

How do you work out any determinant? . . . Just crank it out!

However, we can use some Properties on this one . . .

Here's an important property of determinants.

If a multiple of one row is added to (subtracted from) another row,
. . the value of the determinant is unchanged.

Suppose we have: .$\displaystyle \begin{vmatrix}a&b&c \\ d&e&f \\ g&h&1\end{vmatrix}$. . Its value is:.$\displaystyle (aei + bfg + cdh) - (afh + bdi + ceg)$ .[1]

Subtract row-2 minus row-1: .$\displaystyle \begin{vmatrix}a&b&c \\ d-a & e-b&f-c \\ g&h&i \end{vmatrix}$

Subtract row-3 minus twice row-1: .$\displaystyle \begin{vmatrix}a&b&c \\ d-a & e-b&f-c \\ g-2a & h-2b & i-2c \end{vmatrix}$

This determinant has the same value . . . the value at [1].

11. ## Re: Determinants of this matrix - pls help

Is the number one that you put in the bottom right hand corner of the determinant delibrate?

Suppose we have: $\begin{vmatrix}a&b&c \\ d&e&f \\ g&h&1\end{vmatrix}$ . . Its value is:. (aei + bfg + cdh) - (afh + bdi + ceg) .[1]

And is the determinant alpha once again?

12. ## Re: Determinants of this matrix - pls help

another way to see that subtracting 3 times row 1 from row 2 does not change the determinant:

let P be the matrix:

$\displaystyle P = \begin{bmatrix}1&0&0\\-3&1&0\\0&0&1 \end{bmatrix}$

then

$\displaystyle PA = \begin{bmatrix}a&b&c\\d-3a&e-3b&f-3c\\g&h&i \end{bmatrix}$

(verfiy this!)

since det(PA) = det(P)det(A), to find det(PA), we need to calculate det(P).

det(P) = (1)(1)(1) + (0)(0)(0) + (0)(-3)(0) - (0)(1)(0) - (0)(0)(1) - (1)(0)(-3) = 1 + 0 + 0 - 0 - 0 - 0 = 1

therefore, det(PA) = det(A) = α

(this was what Goku was hinting at his post).

you can see that the value of det(P) won't change if we replace "-3" by "r", and furthermore, it really doesn't matter "where" r goes (as long as it's not on the main diagonal), because there will always be some other 0 on any diagonal (or "extended diagonal", like when you "loop around") containing r to "cancel it out". so subtracting r times any row from a DIFFERENT row, will never change the determinant of a 3x3 matrix.

13. ## Re: Determinants of this matrix - pls help

Thank you very much for the explanation